Question

In Young's double slit experiment using sodium light $\left( {\lambda = 5898\;\mathop A\limits^ \circ } \right)$ 92 fringes are seen. If given colour $\left( {\lambda = 5461\;\mathop A\limits^ \circ } \right)$ is used, how many fringes will be seen
A. 62
B. 67
C. 85
D. 99

Answer 1

The above problem can be resolved by undertaking the condition that n fringes will form when the magnitude of the incident light wavelength changes to some new value. Then the critical situation is applied by the relation of the number of fringes formed newly and the number of fringes formed in the previous case, which is at the light wavelength's initial magnitude.

Complete step by step answer:
Let the length of the screen on which the fringes can be observed be $L$.
Thus, the fringe width obtained, if 92 fringes are formed. The mathematical relation for the given condition is,
${L_1} = \dfrac{L}{{92}}$
As, the mathematical expression for the fringe width is, $w = \dfrac{{D\lambda }}{d}$.
Here, d is the distance between the slit and D is the distance between the screen and the slit.
Hence, for the new wavelength $\left( {{\lambda _1}} \right)$. Let there be no fringes.
Then the mathematical relation obtained for the given condition is,
$n = 92 \times \dfrac{\lambda }{{{\lambda _1}}}$
Solving the above equation as,

n = 92 \times \dfrac{\lambda }{{{\lambda _1}}}\\\ n = 92 \times \dfrac{{5898\;\mathop A\limits^ \circ }}{{5461\;\mathop A\limits^ \circ }}\\\ n = 99 \end{array}$$ Therefore, the number of fringes seen is 99 **So, the correct answer is “Option D”.** **Note:** To resolve the given problem, one must be aware of the fundamentals of Young's double slit experiment and the terms related to the concept of Young's double-slit experiment. The terms like path difference, phase difference, and the angle with which light rays fall on the screen and the constants like slit width and distance between the screen and the slit are also required to remember.